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A solenoid 86.0 cm long has a radius of 2.40 cm and a winding of 1000 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Respuesta :

leena

Hello!

We can use the following equation:

[tex]B = \mu_0 ni[/tex]

B = Magnetic field strength (T)
μ₀ = Permeability of free space (Tm/A)

n = # of loops per unit length

We are given the total number of loops (N), so divide by length:
[tex]n = \frac{N}{l} = \frac{1000}{.86} = 1162.79[/tex]

Now, plug this value into the equation for the magnetic field strength:
[tex]B = (4\pi \times 10^{-7})(1162.79)(3.6) = \boxed{0.00526 T}[/tex]